I'm reading a book on nonlocal diffusion problems and it claims the only steady-state solution of the following nonlocal problem is $u \equiv 0$:
\begin{equation*} \left\{ \begin{array}{ll} u_t(x,t) = \int_{\mathbb{R}^n} J(x-y) u(y,t)dy - u(x,t), & x \in \Omega, t > 0 \\ u(x,t) = 0, & x \notin \Omega, t > 0 \\ u(x,0) = u_0(x), & x \in \Omega \end{array} \right. \end{equation*} where J is continuous, nonnegative, radial, $J(0) > 0$ and $\int_{\mathbb{R}^n} J(x) dx = 1$. A steady-state solution clearly satisfies \begin{equation*} 0 = \int_{\mathbb{R}^n} J(x-y) u(y,t) dy - u(x,t) \end{equation*} for any $x \in \Omega$. The book claims that $u = 0$ in $\mathbb{R}^n \backslash \Omega$ implies \begin{equation*} u(x) = \int_{\mathbb{R}^n} J(x-y) u(y) dy \end{equation*} for any $x \in \mathbb{R}^n$. How are they able to extend the result from $x \in \Omega$ to $x \in \mathbb{R}^n$? Once they obtain this result they say it clearly implies that $u \equiv 0$ which I am not seeing. If I assume the initial data is nonnegative, it seems trivial, but they make no such assumption. Any help would be greatly appreciated.
Take Fourier transform, we have $\hat{u}(\xi) = \hat{J}(\xi)\hat{u}(\xi)$, so the support of $\hat{u}$ is contained in where$\hat{J}=1$, but since $J$ is positive and $\int J=1$, we know $\hat{J}=1$ at exactly $\xi = 0$ (see this post Supremum of absolute value of the Fourier transform equals $1$, and it is attained exactly at $0$)
so the support of $\hat{u}$ is $\{0\}$, and we know $u$ as a termpered distribution must be a polynomial, but $u = 0$ for $x \notin \Omega$, hence $u$ must be $0$ on all of $\mathbb{R}^n$.